estimating absolute value of complex function Let $f(z) = \frac{1}{z^2 - 2z + 4 }$  where $|z| = R > 1 $. How can I estimate $|f(z)|$ to get an upper bound of the form $\frac{ M}{|z|^p }$ where $M > 0$ and $p>1$ ? I know 
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|z^2 - 2z|} = \frac{1}{|z||z-2|}$$
and here is where I got Stuck. Any ideas?
 A: You can't get such an estimate valid for all $|z| > 1$ (since $f$ has poles at $z=1\pm i\sqrt{3}$ and $|1\pm i\sqrt{3}|=2$.
On the other hand you probably only need the estimate for sufficiently large $|z|$, and then you have
$$
\left| \frac{1}{z^2-2z+4} \right| \le \frac{1}{|z|^2-2|z|-4} \le \frac{1}{R^2 - 2R - 4} \le \frac{1}{R^2 - \frac14 R^2 - \frac 14 R^2} = \frac{2}{R^2}
$$
for $|z| \ge R$ as long as $\frac12 R^2 \ge 2R$ and $\frac14 R^2 \ge 4$. (I.e. if $R \ge 4$.) With more careful estimates you can get $R$ as close to $2$ as you like, but that's almost certainly not needed for what you want to do.)
A: Clearly $|z^2 -2z+4|=|z^2-2(z-2)|=|2(z-2)-z^2|\geq ||2z-4|-|z^2||$.
$\Rightarrow$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{||2z-4|-|z^2| |}$$
So by triangle inequality we have that  $|2z-4|\geq |2z|-|4|$
Since $|z|=R>1$ , there exists $a \in \mathbb R^+$ such that $4=a|z|$ and $a>1$ .
Therefore $$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{||2z|-|4|-|z^2| |}$$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{\left | |2z|-a|z|-|z^2| \right |}$$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|(2-a)|z|-|z^2|| }$$
So $$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|(2-a)|z|-|z^2|| }=\frac{1}{||z^2|-(2-a)|z|| }$$
Since $|z|>1$ , $|z|^2 >|z|$.
So $||z^2|-(2-a)|z|| \geq ||z^2|-(2-a)|z^2||=||z^2|(a-1)||$
Now put $M= \frac{1}{a-1} >0$ as $a>1$ .
Then $$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{M}{|z|^2 }$$
A: Take $z=\rho e^{i\theta}$ and $z_0=ae^{i\theta_0}$, then
$$
|z-z_0|\geq||z|-|z_0||=|\rho-a|=\rho-a,
$$
the last passage for $\rho>a$.
Then
$$
\left|\frac{1}{z-z_0}\right|\leq\frac{1}{\rho-a}\leq\frac{M}{\rho}
$$
the last inequality holds if $\rho\geq a+\displaystyle\frac{a}{M-1}$ and $M>1.$
So
$$
\left|\frac{1}{z^2-2z+4}\right|\leq\frac{1}{\rho-a}\cdot\frac{1}{\rho-b}\leq\frac{M}{\rho^2}
$$
for $\rho\geq\max\left\{a+\displaystyle\frac{a}{M_1-1},b+\displaystyle\frac{b}{M_2-1}\right\}$, $M_1>1,M_2>1$ and $M=M_1M_2$.

In general, if $P$ is a polynomial of degree $n$, with leading coefficient $a_n$, and $R$ is the maximum of the modulus of its roots, then
$$
\left|\frac{1}{P(z)}\right|\leq\frac{2^n}{|a_n||z|^n},\qquad\forall|z|\geq 2R
$$
(this is obtained from the previous analysis taking $M_i=2, \forall i$).
